Let non-homogeneous system $Ax=b$ nas solution $$x=\begin{bmatrix} 1 \\ 0 \\1 \\ 0\\ \end{bmatrix} + \alpha_1\begin{bmatrix} 1\\ 1\\ -1 \\ 0\\ \end{bmatrix} + \alpha_2\begin{bmatrix} 1\\ 0\\ 1 \\ 1\\ \end{bmatrix}; \space \alpha_1 , \alpha_2 \in \Bbb{R} \\ $$ a)A is an m × n matrix of rank r. Describe all possible values of m, n, and r. $$ $$b) Find matrix A if $b=\begin{bmatrix} 1 \\ 2 \\ 1 \\ \end{bmatrix} \\$ $$-$$ a)I found a solution that matrix $A$ has 4 columns (why exactly 4??) , and because homogeneous system result are vectors $\begin{bmatrix} 1\\ 1\\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1\\ 0\\ 1 \\ 1\\ \end{bmatrix}$ they form null space , so $dim Ker(A)=2$ and from rank-nullity theorem we have that r=2 so then $m \ge r$. I don't know how can I find from given solution $x$ how many columns matrix $A$ has and why number of rows is $m\ge r$ ?
b)If my matrix $A$ has 4 coumns and $rankA=2$ I have to write 2 independent vectors and two non-independent.How to know which two vectors I should use to create column space of $A$ if my $b=\begin{bmatrix} 1 \\ 2 \\ 1 \\ \end{bmatrix} \\$ , and it's obviously that other two have to be linear combination of them to form null space. $$ $$ I found that solution for b) is $$A=\begin{bmatrix} 1 & -1 & 0 &-1\\ 1 & 0 & 1&-2 \\ 1 & -1 & 0 &-1\\ \end{bmatrix}$$ First and third column are lineary independent and their sum gives me a vector b, does it mean that linear combination of independent vectors gives me a vector b from $Ax=b$ ?
First of all $Ax=b$ with $b\in\mathbb R^3$ and $x\in\mathbb R^4$ means that $A$ is a $3\times 4$ matrix. This does not make sense otherwise. Write the components of $A$ as $a_{ij}$ for $i=1,\dots,3$ and $j=1,\dots,4$. The equation $Ax=b$ for those specific values of $x$ and $b$ give $$ (Ax)_i=a_{i1}+a_{i3}+\alpha_1 (a_{i1}+a_{i2}-a_{i3})+\alpha_2(a_{i1}+a_{i3}+a_{i4})=b_i. $$ Since this is true for all $\alpha_1$ and $\alpha_2$ in $\mathbb R$, we need the combinations in the brackets to vanish. So we have for $i=1,\dots,3$ $$ a_{i1}+a_{i3}=b_i,\quad a_{i1}+a_{i2}-a_{i3}=0,\quad a_{i1}+a_{i3}+a_{i4}=0. $$ It is not too hard to see that we then have a three dimensional space of solutions spanned by the values of $a_{11},a_{21}$ and $a_{31}$ as $$ A=\begin{pmatrix}a_{11}&b_1-2a_{11}&b_1-a_{11}&-b_1\\a_{21}&b_2-2a_{21}&b_2-a_{21}&-b_2\\a_{31}&b_3-2a_{31}&b_3-a_{31}&-b_3\end{pmatrix}. $$